codeforces301A. Yaroslav and Sequence_yaroslav has an array, consisting of (2·n

本文探讨了Yaroslav面对包含(2·n - 1)个整数的序列，通过改变其中n个元素的符号来最大化序列总和的问题。介绍了如何根据序列中负数的数量确定最优策略，并提供了具体的实现代码。

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Yaroslav has an array, consisting of (2·n - 1) integers. In a single operation Yaroslav can change the sign of exactly n elements in the array. In other words, in one operation Yaroslav can select exactly n array elements, and multiply each of them by -1.

Yaroslav is now wondering: what maximum sum of array elements can be obtained if it is allowed to perform any number of described operations?

Help Yaroslav.

Input

The first line contains an integer n (2 ≤ n ≤ 100). The second line contains (2·n - 1) integers — the array elements. The array elements do not exceed 1000 in their absolute value.

Output

In a single line print the answer to the problem — the maximum sum that Yaroslav can get.

Examples

input
2
50 50 50

output
150

input
2
-1 -100 -1

output
100

Note

In the first sample you do not need to change anything. The sum of elements equals 150.

In the second sample you need to change the sign of the first two elements. Then we get the sum of the elements equal to 100.

找规律，刚开始做此题是看成要连续的n个数读错题意，然后各种yy就是没思路，然后在C++课上想终于想到了思路，考虑数组中负数的个数即可。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=210;
int num[maxn];
bool cmp(int a,int b){
	return abs(a)<abs(b);
}
int main()
{
	int n,i,j,k;
	while(scanf("%d",&n)!=EOF){
		int cnt=0,ans=0;
		for(i=0;i<2*n-1;++i){
			scanf("%d",&num[i]);
			if(num[i]<0)cnt++;
			ans=ans+abs(num[i]);
		}
		sort(num,num+2*n-1,cmp);
		if(cnt%2==0||cnt==n||abs(n-cnt)%2==0){
			printf("%d\n",ans);
		}
		else {
			printf("%d\n",ans-2*abs(num[0]));
		}
	}
	return 0;
}